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Math Help - Another Trig Problem

  1. #1
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    Another Trig Problem

    If your not too busy...

    1-tan^2X =1-2tan^2X
    1+tan^2X
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  2. #2
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    Quote Originally Posted by Dominic O'L View Post
    If your not too busy...

    1-tan^2X =1-2tan^2X
    1+tan^2X
    What is the question? Solve for x ....?
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  3. #3
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    No, just prove that they are equal.
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  4. #4
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    Quote Originally Posted by Dominic O'L View Post
    No, just prove that they are equal.
    They're not. eg. Does it work for x = \frac{\pi}{4} ....?
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  5. #5
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    oh sorry i read the question incorrectly

    1-tan^2X =1-2sin^2X
    1+tan^2X
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  6. #6
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    Quote Originally Posted by Dominic O'L View Post
    oh sorry i read the question incorrectly

    1-tan^2X =1-2sin^2X
    1+tan^2X
    First, use long division on the left hand side.

    You should find

    \frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\tan^2{x} + 1}

    Remember that \tan^2{x} + 1 = \sec^2{x}

    So \frac{1 - \tan^2{x}}{1 + \tan^2{x}} = -1 + \frac{2}{\sec^2{x}}

     = -1 + 2\cos^2{x}

     = -1 + 2(1 - \sin^2{x}), since \cos^2{x} + \sin^2{x} = 1...

     = -1 + 2 - 2\sin^2{x}

     = 1 - 2\sin^2{x}.
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  7. #7
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    Sorrry, but I think I'm supposed to solve it without sec.
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  8. #8
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    Quote Originally Posted by Dominic O'L View Post
    Sorrry, but I think I'm supposed to solve it without sec.
    Then write it as

    -1 + \frac{2}{\frac{1}{\cos^2{x}}}.

    Same thing.
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  9. #9
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    Hello Dominc O'L

    \frac{1-\tan^2x}{1+\tan^2x}

    =\frac{1-\tan^2x}{1+\tan^2x}\cdot \frac{\cos^2x}{\cos^2x}

    =\frac{\cos^2x - \sin^2x}{\cos^2x + \sin^2x}

    = \frac{(1-\sin^2x)-\sin^2x}{1}

    = 1 - 2\sin^2x

    Grandad
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